ml aggarwal solutions class 10 maths chapter 21: measures

ML Aggarwal Solutions Class 10 Maths Chapter 21: Measures of Central Tendency

Exercise 21.1

1. (a) Calculate the arithmetic mean of 5.7, 6.6, 7.2, 9.3, 6.2.(b) The weights (in kg) of 8 new born babies are 3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2. Find the mean weight of thebabies.

Solution: (a) Given observations are 5.7, 6.6, 7.2, 9.3, 6.2.Number of observations = 5Mean = sum of observations / number of observations∴Mean = (5.7+6.6+7.2+9.3+6.2)/5= 35/5 = 7Hence the mean of the given observations is 7.

(b) Given weight of babies are 3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2Number of observations = 8Mean = sum of observations / number of observations∴Mean = (3+3.2+3.4+3.5+4+3.6+4.1+3.2)/8= 28/8 = 3.5 kgHence the mean of the weight of babies is 3.5 kg.

2.The marks obtained by 15 students in a class test are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20find(i) the mean of their marks.(ii) the mean of their marks when the marks of each student are increased by 4.(iii) the mean of their marks when 2 marks are deducted from the marks of each student.(iv) the mean of their marks when the marks of each student are doubled.

Solution: (i) Marks obtained by students are 12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20.Number of students = 15Mean = sum of observations / number of observations= 12+14+07+09+23+11+08+13+11+19+16+24+17+03+20= 207/15= 13.8Hence the mean of their marks is 13.8.

(ii) If mark of each student is increased by 4, total increased marks = 4×15 = 60Total increase in sum of marks = 207+60 = 267∴ mean = sum of marks/number of students∴mean = 267/15 = 17.8Hence the mean is 17.8.

(iii) If mark of each student is deducted by 2, total deducted marks = 2×15 = 30Total decrease in sum of marks = 207-30 = 177∴mean = sum of marks/number of students∴mean = 177/15 = 11.8

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ML Aggarwal Solutions Class 10 Maths Chapter : Measures of Central Tendency

Hence the mean is 11.8. (iv) If mark of each student is doubled , then new sum of marks = 2×207 = 414 ∴mean = new sum of marks/number of students ∴mean = 414/15 = 27.6 Hence the mean is 27.6. 3. (a) The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x. (b) The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15 find the 9th variate. Solution: (a)Given observations are 6, y, 7, x, 14. Mean = 8 Number of observations = 5 Mean = Sum of observations/number of observations ∴8 = (6+y+7+x+14)/5 ∴40 = 27+x+y 40-27 = x+y 13 = x+y ∴y = 13-x Hence the answer is y = 13-x. (b)Given mean = 11 Number of variates = 9 Variates are 7, 12, 9, 14, 21, 3, 8 ,15 Let the 9th variate be x. Sum of variates = 7+12+9+14+21+3+8+15+x = 89+x Mean = Sum of variates/number of variates ∴11 = (89+x)/9 11×9 = 89+x 99 = 89+x ∴x = 99-89 = 10 Hence the 9th variate is 10. 4. (a) The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes

years. What is the age of the girl ? (b) In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that marks of one student was wrongly copied as 21 instead of 29. Find the correct mean. Solution: (a)Given mean age = 13 Number of students = 33 Sum of ages = mean ×number of students = 13×33 = 429

After a girl leaves, the mean of 32 students becomes = 207/16




Now sum of ages = 32×207/16 = 414 So the age of the girl who left = 429-414 = 15 years. Hence the age of the girl who left is 15 years. (b)Mean of marks = 18.2 Number of students = 40 Total marks of 40 students = 40×18.2 = 728 Difference of marks when copied wrongly = 29-21 = 8 So total marks = 728+8 = 736 ∴mean = 736/40 = 18.4 Hence the correct mean is 18.4. 5. Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of the remaining numbers is 18. Solution: Mean of 10 numbers = 13 ∴Sum of numbers = 13×10 = 130 Mean of remaining 15 numbers = 18 ∴Sum of numbers = 15×18 = 270 ∴Sum of all numbers = 130+270 = 400 ∴Mean = sum of numbers/25 = 400/25 = 16 Hence the mean of 25 numbers is 16. 6. Find the mean of the following distribution: Number 5 10 15 20 25 30 35 Frequency 1 2 5 6 3 2 1

Solution: Number (x) Frequency (f) fx

5 1 5×1 = 5

10 2 10×2 = 20

15 5 15×5 = 75

20 6 20×6 = 120

25 3 25×3 = 75

30 2 30×2 = 60

35 1 35×1 = 35

Total Ʃf = 20 Ʃfx = 390

∴Mean = Ʃfx/Ʃf




= 390/20 = 19.5 Hence the mean is 19.5. 7. The contents of 100 match boxes were checked to determine the number of matches they contained No. of matches

35 36 37 38 39 40 41

No. of boxes

6 10 18 25 21 12 8

(i) Calculate, correct to one decimal place, the mean number of matches per box. (ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to; bring the mean up to exactly 39 matches. (1997) Solution: (i) No. of matches (x) Number of boxes (f) fx 35 6 35×6 = 210 36 10 36×10 = 360 37 18 37×18 = 666 38 25 38×25 = 950 39 21 39×21 = 819 40 12 40×12 = 480 41 8 41×8 = 328 Total Ʃf = 100 Ʃfx = 3813 Mean = Ʃfx/Ʃf = 3813/100 = 38.13 = 38.1 Hence the mean is 38.1. (ii)New mean = 39 ∴Ʃfx = 39×100 = 3900 So number of extra matches to be added = 3900-3813 = 87 Hence the number of extra matches to be added is 87. 8. Calculate the mean for the following distribution : Pocket money (in Rs)

60 70 80 90 100 110 120

No. of students

2 6 13 22 24 10 3

Solution:




Pocket money in Rs (x)

Number of students (f)

fx

60 2 60×2 = 120 70 6 70×6 = 420 80 13 80×13 = 1040 90 22 90×22 = 1980 100 24 100×24 = 2400 110 10 110×10 = 1100 120 3 120×3 = 360 Total Ʃf = 80 Ʃfx = 7420 Mean = Ʃfx/Ʃf = 7420/80 = 92.75 Hence the mean is 92.75. 9. Six coins were tossed 1000 times, and at each toss the number of heads were counted and the results were recorded as under : No. of heads 6 5 4 3 2 1 0

No. of tosses 20 25 160 283 338 140 34

Calculate the mean for this distribution. Solution: No. of heads (x) No. of tosses (f) fx 6 20 6×20 = 120 5 25 5×25 = 125 4 160 4×160 = 640 3 283 3×283 = 849 2 338 2×338 = 676 1 140 1×140 =140 0 34 0×34 = 0 Total Ʃf = 1000 Ʃfx = 2550 Mean = Ʃfx/Ʃf = 2550/1000 = 2.55 Hence the mean is 2.55. 10. Find the mean for the following distribution. Numbers 60 61 62 63 64 65 66




Cumulative frequency

8 18 33 40 49 55 60

Solution: Numbers (x)

Cumulative frequency

Frequency (f) fx

60 8 8 60×8 = 480

61 18 18-8 = 10 61×10 = 610

62 33 33-18 = 15 62×15 = 930 63 40 40-33 = 7 63×7 = 441

64 49 49-40 = 9 64×9 = 576

65 55 55-49 = 6 65×6 = 390 66 60 60-55 = 5 66×5 = 330

Total Ʃf = 60 Ʃfx = 3757

∴Mean = Ʃfx/Ʃf = 3757/60 = 62.616 = 62.62 Hence the mean is 62.62. 11. Category A B C D E F G Wages (in Rs) per day

50 60 70 80 90 100 110

No. of workers

2 4 8 12 10 6 8

(i) Calculate the mean wage correct to the nearest rupee (1995) (ii) If the number of workers in each category is doubled, what would be the new mean wage ? Solution: Category Wages in Rs. (x) No. of workers f fx A 50 2 100 B 60 4 240 C 70 8 560 D 80 12 960 E 90 10 900 F 100 6 600 G 110 8 880 Total Ʃf = 50 Ʃfx = 4240 ∴Mean = Ʃfx/Ʃf = 4240/50




= 84.8 = 85 Hence the mean is 85. (ii)If number of workers is doubled, then total number of workers = 50×2 = 100 So wages will be doubled. ∴Total wages = 4240×2 = 8480 ∴Mean = Ʃfx/Ʃf = 8480/100 = 84.8 = 85 Hence the mean is 85. 12.If the mean of the following distribution is 7.5, find the missing frequency ” f “. Variate 5 6 7 8 9 10 11 12

Frequency 20 17 f 10 8 6 7 6

Solution: Variate (x) Frequency (f) fx 5 20 100 6 17 102 7 f 7f 8 10 80 9 8 72 10 6 60 11 7 77 12 6 72 Total Ʃf = 74+f Ʃfx = 563+7f Given mean = 7.5 ∴Mean = Ʃfx/Ʃf 7.5= (563+f)/(74+7f) 7.5×(74+f) = 563+7f 555+7.5f = 563+7f 7.5f-7f = 563-555 0.5f = 8 ∴f = 8/0.5 = 16 Hence the value of missing frequency f is 16. 13. Find the value of the missing variate for the following distribution whose mean is 10 Variate (xi)

5 7 9 11 _ 15 20

Frequency (fi)

4 4 4 7 3 2 1




Solution: Let the missing variate be x. Variate (xi) Frequency (fi) fixi 5 4 20 7 4 28 9 4 36 11 7 77 x 3 3x 15 2 30 20 1 20 Total Ʃfi =25 Ʃfixi = 211+3x Given mean = 10 Mean = Ʃfixi/Ʃfi

∴10= (211+3x)/25 10×25 = 211+3x 250 = 211+3x 250-211 = 3x 39 = 3x x = 39/3 = 13 Hence the missing variate is 13. 14. Marks obtained by 40 students in a short assessment are given below, where a and b are two missing data. Marks 5 6 7 8 9

No. of students

6 a 16 13 b

If the mean of the distribution is 7.2, find a and b. Solution: Marks (x) No. of students (f) fx 5 6 30 6 a 6a 7 16 112 8 13 104 9 b 9b Total Ʃf = 35+a+b Ʃfx = 246+6a+9b Given number of students = 40 ∴ Ʃf = 35+a+b = 40 a+b = 40-35 = 5 a = 5-b ……(i) Mean = Ʃfx/Ʃf

Given mean = 7.2 ∴( 246+6a+9b) /40 = 7.2




∴( 246+6a+9b) = 40×7.2 ∴( 246+6a+9b) = 288 ∴6a+9b = 288-246 ∴6a+9b = 288-246 ∴6a+9b = 42 2a+3b = 14 …..(ii) Substitute (i) in (ii) 2(5-b)+3b = 14 10-2b+3b = 14 10+b = 14 ∴b = 14-10 = 4 ∴a = 5-4 = 1 Hence the value of a and b is 1 and 4 respectively. 15. Find the mean of the following distribution. Class interval

0-10 10-20 20-30 30-40 40-50

Frequency 10 6 8 12 5 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi 0-10 10 5 50 10-20 6 15 90 20-30 8 25 200 30-40 12 35 420 40-50 5 45 225 Total Ʃfi = 41 Ʃfixi = 985 Mean = Ʃfixi/ Ʃfi

= 985/41 = 24.024 = 24.02 (approx) Hence the mean of the distribution is 24.02. 16. Calculate the mean of the following distribution: Class interval

0-10 10-20 20-30 30-40 40-50 50-60

Frequency 8 5 12 35 24 16

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi




0-10 8 5 40 10-20 5 15 75 20-30 12 25 300 30-40 35 35 1225 40-50 24 45 1080 50-60 16 55 880 Total Ʃfi = 100 Ʃfixi = 3600 Mean = Ʃfixi/ Ʃfi

= 3600/100 = 36 Hence the mean of the distribution is 36. 17. Calculate the mean of the following distribution using step deviation method: Marks 0-10 10-20 20-30 30-40 40-50 50-60

No. of students

10 9 25 30 16 10

Solution: Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 25 Class size (h) = 10

Class Interval No. of students (fi) Class mark (xi) di = xi - A ui = di/h fiui 0-10 10 5 -20 -2 -20 10-20 9 15 -10 -1 -9 20-30 25 25 0 0 0 30-40 30 35 10 1 30 40-50 16 45 20 2 32 50-60 10 55 30 3 30

Total ∑fi = 100 ∑fiui =63

By step deviation method, Mean = x̄ = A+h∑fiui /∑fi

= 25+10(63/100) = 25+10×0.63 = 25+6.3 =31.3 Hence the mean of the distribution is 31.3. 18. Find the mean of the following frequency distribution: Class intervals

0-50 50-100 100-150 150-200 200-250 250-300

frequency 4 8 16 13 6 3




Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi 0-50 4 25 100 50-100 8 75 600 100-150 16 125 2000 150-200 13 175 2275 200-250 6 225 1350 250-300 3 275 825 Total Ʃfi = 50 Ʃfixi = 7150 Mean = Ʃfixi/ Ʃfi

= 7150/50 =143 Hence the mean of the distribution is 143. 19. The following table gives the daily wages of workers in a factory: Wages in Rs.

45-50 50-55 55-60 60-65 65-70 70-75 75-80

No. of workers

5 8 30 25 14 12 6

Calculate their mean by short cut method. Solution: Class mark, xi = (upper class limit + lower class limit)/2 Assumed mean, A = 62.5

Wages in Rs. No. of workers (fi) Class mark (xi) di = xi - A fidi 45-50 5 47.5 -15 -75 50-55 8 52.5 -10 -80 55-60 30 57.5 -5 -150 60-65 25 62.5 0 0 65-70 14 67.5 5 70 70-75 12 72.5 10 120 75-80 6 77.5 15 90

Total ∑fi = 100 ∑fidi = -25

By short cut method, Mean = x̄ = A+∑fidi /∑fi = 62.5+-25/100 = 62.5-0.25 = 62.25 Hence the mean of the distribution is Rs.62.25. 20. Calculate the mean of the distribution given below using the short cut method.




Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80

No. of students

2 6 10 12 9 7 4

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Assumed mean, A = 45.5

Marks No. of students (fi) Class mark (xi) di = xi - A fidi 11-20 2 15.5 -30 -60 21-30 6 25.5 -20 -120 31-40 10 35.5 -10 -100 41-50 12 45.5 0 0 51-60 9 55.5 10 90 61-70 7 65.5 20 140 71-80 4 75.5 30 120

Total ∑fi = 50 ∑fidi = 70

By short cut method, Mean = x̄ = A+∑fidi /∑fi = 45.5+70/50 = 45.5+1.4 = 46.9 Hence the mean of the distribution is Rs.46.9. 21. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. No. of days

0-6 6-10 10-14 14-20 20-28 28-38 38-40

No. of students

11 10 7 4 4 3 1

Solution: Class mark, xi = (upper class limit + lower class limit)/2 No. of days Frequency fi Class mark xi fixi 0-6 11 3 33 6-10 10 8 80 10-14 7 12 84 14-20 4 17 68 20-28 4 24 96 28-38 3 33 99 38-40 1 39 39 Total Ʃfi = 40 Ʃfixi = 499




Mean = Ʃfixi/ Ʃfi

= 499/40 =12.475 Hence the mean number of days a student was absent is 12.475. 22. The mean of the following distribution is 23.4. Find the value of p. Class intervals

0-8 8-16 16-24 24-32 32-40 40-48

Frequency 5 3 10 P 4 2

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-8 5 4 20 8-16 3 12 36 16-24 10 20 200 24-32 P 28 28P 32-40 4 36 144 40-48 2 44 88 Total Ʃfi = 24+P Ʃfixi = 488+28P Given mean = 23.4 ∴Mean = Ʃfixi/ Ʃfi

∴23.4 = (488+28P)/(24+P) ∴23.4×(24+P) = 488+28P ∴561.6+23.4P = 488+28P ∴561.6-488 = 28P -23.4P 73.6 = 4.6 P ∴P = 73.6/4.6 = 16 Hence the value of P is 16. 23. The following distribution shows the daily pocket allowance for children of a locality. The mean pocket allowance is Rs. 18. Find the value of f. Daily pocket allowance in Rs.

11-13 13-15 15-17 17-19 19-21 21-23 23-25

No. of children

3 6 9 13 f 5 4

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Daily pocket No. of children fi Class mark xi fixi




allowance in Rs. 11-13 3 12 36 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 f 20 20f 21-23 5 22 110 23-25 4 24 96 Total Ʃfi = 40+f Ʃfixi = 704+20f Given mean = 18 ∴Mean = Ʃfixi/ Ʃfi

∴18 = (704+20f)/( 40+f) ∴18×(40+f) = 704+20f ∴720 +18f = 704+20f 720-704 = 20f-18f 16 = 2f f = 16/2 = 8 Hence the value of f is 8. 24. The mean of the following distribution is 50 and the sum of all the frequencies is 120. Find the values of p and q. Class intervals

0-20 20-40 40-60 60-80 80-100

Frequency 17 P 32 q 19

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi

0-20 17 10 170 20-40 P 30 30P 40-60 32 50 1600 60-80 q 70 70q 80-100 19 90 1710 Total Ʃ fi = 68+P+q Ʃfixi =

3480+30P+70q Given sum of all frequencies, Ʃ fi = 120 ∴68+P+q = 120 ∴P+q = 120-68 = 52 ∴P+q = 52 ∴P = 52-q …(i)




Given mean = 50 ∴Mean = Ʃfixi / Ʃ fi ∴50 = (3480+30P+70q)/120 ∴50×120 = 3480+30P+70q ∴6000 = 3480+30P+70q ∴6000- 3480 = 30P+70q ∴2520 = 30P+70q ∴252 = 3P+7q …(ii) Substitute (i) in (ii) 252 = 3(52-q)+7q 252 = 156-3q+7q 252-156 = 4q 4q = 96 q = 96/4 = 24 ∴P = 52-24 = 28 Hence the value of P and q is 28 and 24 respectively. 25.The mean of the following frequency distribution is 57.6 and the sum of all the frequencies is 50. Find the values of p and q. Class intervals

0-20 20-40 40-60 60-80 80-100 100-120

Frequency 7 P 12 q 8 5

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi

0-20 7 10 70 20-40 P 30 30P 40-60 12 50 600 60-80 q 70 70q 80-100 8 90 720 100-120 5 110 550 Total Ʃ fi = 32+P+q Ʃfixi =

1940+30P+70q Given sum of all frequencies, Ʃ fi = 50 ∴32+P+q = 50 ∴P+q = 50-32= 18 ∴P+q = 18 ∴P = 18-q …(i) Given mean = 57.6 ∴Mean = Ʃfixi / Ʃ fi




∴57.6 = (1940+30P+70q)/50 ∴57.6 ×50 = 1940+30P+70q ∴2880= 1940+30P+70q ∴2880- 1940 = 30P+70q ∴940 = 30P+70q ∴94 = 3P+7q …(ii) Substitute (i) in (ii) 94 = 3(18-q)+7q 94 = 54-3q+7q 94-54= 4q 40 = 4q q = 40/4 = 10 ∴P = 18-10 = 8 Hence the value of P and q is 8 and 10 respectively. 26. The following table gives the life time in days of 100 electricity tubes of a certain make : Life time in days No. of tubes Less than 50 8 Less than 100 23 Less than 150 55 Less than 200 81 Less than 250 93 Less than 300 100 Find the mean life time of electricity tubes. Solution: Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 175 Class size (h) = 50

Class Interval No. of tubes (cf) Class mark (xi) di = xi - A ui = di/h Frequency (fi)

fiui

0-50 8 25 -150 -3 8 -24 50-100 23 75 -100 -2 15 -30 100-150 55 125 -50 -1 32 -32 150-200 81 175 0 0 26 0 200-250 93 225 50 1 12 12 250-300 100 275 100 2 7 14

Total ∑fi = 100 ∑fiui =-60


= 175+50(-60/100)




= 175+50×-0.60 = 175-30 = 145 Hence the mean of the electricity tubes is 145. 27. Using the information given in the adjoining histogram, calculate the mean correct to one decimal place.

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Assumed mean, A = 45

Class interval frequency (fi) Class mark (xi) di = xi - A fidi 20-30 3 25 -20 -60 30-40 5 35 -10 -50 40-50 12 45 0 0 50-60 9 55 10 90 60-70 4 65 20 80

∑fi = 33 ∑fidi = 60

By short cut method, Mean = x̄ = A+∑fidi /∑fi = 45+60/33




= 45+1.81 = 46.81 = 46.8 [corrected to one decimal place] Hence the mean is Rs.46.8.




Exercise 21.2 1. A student scored the following marks in 11 questions of a question paper : 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7 Find the median marks. Solution: Arranging the data in the ascending order 1,2,2,3,4,5,5,6,7,7,8 Here number of terms, n = 11 Here n is odd. So median = [(n+1)/2 ]th observation = (11+1)/2 = 12/2 = 6th observation Here 6th observation is 5. Hence the median is 5. 2. (a) Find the median of the following set of numbers : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 (1990) (b)For the following set of numbers, find the median: 10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15. Solution: (a) Arranging the numbers in ascending order : 0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9 Here, n = 12 which is even Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (12/2 th term + ((12/2)+1)th term) = ½ (6 th term + (6+1)th term) = ½ (6 th term + 7th term) = ½ (4+5) = 9/2 = 4.5 Hence the median is 4.5. (b) Arranging the numbers in ascending order : 3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81 Here, n = 12 which is even Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (12/2 th term + ((12/2)+1)th term) = ½ (6 th term + (6+1)th term) = ½ (6 th term + 7th term) = ½ (15+17) = ½ ×32 = 16 Hence the median is 16. 3. Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5 Solution:




Arranging the numbers in ascending order : 0, 1, 1, 2, 2, 3, 3, 3, 4, 5 Here, n = 10 which is even Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (10/2 th term + ((10/2)+1)th term) = ½ (5 th term + (5+1)th term) = ½ (5 th term + 6th term) = ½ (2+3) = ½ ×5 = 2.5 Hence the median is 2.5. Mean = sum of the observations/ number of observations = Ʃxi/n = (0+1+1+2+2+3+3+3+4+5)/10 = 24/10 = 2.4 Hence the mean is 2.4. 4.The median of the observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean. Solution: Observation are as follows : 11, 12, 14, (x-2), (x+4), (x+9), 32, 38, 47 n = 9 Here n is odd. So median = ((n+1)/2)th term = (9+1)/2 )th term = 5th term = x+4 Given median = 24 ∴x+4 = 24 ∴x = 24 -4 = 20

Sum of observations = 11+12+14+(x-2)+(x+4)+(x+9)+32+38+47 = 165+3x Substitute x = 20 Sum of observations = 165+3×20 = 165+60 = 225 Mean = Sum of observations /number of observations = 225/9 = 25 Hence the value of x is 20 and mean is 25. 5.The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m-1 and median q. Find (i) p (ii) q (iii) the mean of p and q.




Solution: (i) Mean of 1, 7, 5, 3, 4, 4 is m. Here n = 6 ∴Mean, m = (1+7+5+3+4+4)/6 ∴m = 24/6 ∴m = 4 Given the numbers 3, 2, 4, 2, 3, 3, p have mean m-1. So m-1 = (3+2+4+2+3+3+p)/7 ∴4-1 = (17+p)/7 ∴3 = (17+p)/7 ∴3×7 = 17+p ∴21 = 17+p ∴p = 21-17 ∴p = 4 Hence the value of p is 4. (ii) Given the numbers have median q. Arranging them in ascending order 2, 2, 3, 3, 3, 4, 4 Here n = 7 which is odd So median = ((n+1)/2)th term q = ((7+1)/2 )th term q = (8/2 )th term q = 4th term q = 3 So value of q is 3. (iii)mean of p and q = (p+q)/2 = (4+3)/2 = 7/2 = 3.5 Hence the mean of p and q is 3.5. 6. Find the median for the following distribution: Wages per day in Rs.

38 45 48 55 62 65

No. of workers

14 8 7 10 6 2

Solution: We write the distribution in cumulative frequency table. Wages per day in Rs. No. of workers (f) Cumulative frequency 38 14 14 45 8 22 48 7 29




55 10 39 62 6 45 65 2 47 Here total number of observations, n = 47 which is odd. So median =(( n+1)/2)th term = ((47+1)/2 )th term = (48/2 )th term = 24th term = 48 [Since 23rd to 29th observation is 48] Hence the median is 48. 7. Find the median for the following distribution. Marks 35 45 50 64 70 72

No. of students

3 5 8 10 5 5

Solution: We write the distribution in cumulative frequency table. Marks No. of students (f) Cumulative frequency 35 3 3 45 5 8 50 8 16 64 10 26 70 5 31 72 5 36 Here total number of observations, n = 36 which is even. Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (36/2 th term + ((36/2)+1)th term) = ½ (18 th term + (18+1)th term) = ½ (18 th term + 19th term) = ½ (64+64) [Since 17th to 26th observation is 64] = ½ ×128 = 64 Hence the median is 64. 8.Marks obtained by 70 students are given below : Marks 20 70 50 60 75 90 40

No. of students

8 12 18 6 9 5 12




Calculate the median marks. Solution: We write the marks in ascending order in cumulative frequency table. Marks No. of students (f) Cumulative frequency 20 8 8 40 12 20 50 18 38 60 6 44 70 12 56 75 9 65 90 5 70 Here total number of observations, n = 70 which is even. Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (70/2 th term + ((70/2)+1)th term) = ½ (35 th term + (35+1)th term) = ½ (35 th term + 36th term) = ½ (50+50) [Since all observations from 21st to 38th are 50] = ½ ×100 = 50 Hence the median is 50. 9. Calculate the mean and the median for the following distribution : Number 5 10 15 20 25 30 35

Frequency 1 2 5 6 3 2 1

Solution: We write the numbers in cumulative frequency table. Marks (x) No. of students (f) Cumulative frequency fx 5 1 1 5 10 2 3 20 15 5 8 75 20 6 14 120 25 3 17 75 30 2 19 60 35 1 20 35 Total Ʃf = 20 Ʃfx = 390 Mean = Ʃfx/Ʃf




= 390/20 = 19.5 Hence the mean is 19.5. Here number of observations, n = 20 which is even. So median = = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (20/2 th term + ((20/2)+1)th term) = ½ (10 th term + (10+1)th term) = ½ (10 th term + 11th term) = ½ (20+20) [Since all observations from 9th to 14th are 20] = ½ ×140 = 20 Hence the median is 20. 10. The daily wages in (rupees of) 19 workers are 41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35. find : (i) the median (ii) lower quartile (iii) upper quartile (iv) inter quartile range Solution: Arranging the observations in ascending order 21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53 Here n = 19 which is odd. (i)Median = ((n+1)/2)th term = (19+1)/2 = 20/2 = 10th term = 31 Hence the median is 31. (ii) Lower quartile, Q1 = ((n+1)/4) th term = (19+1)/4 = 20/4 = 5 th term = 27 Hence the lower quartile is 27. (iii)Upper quartile, Q3 = (3(n+1)/4) th term = (3×(19+1)/4) th term = (3×(20/4)) th term = (3×5) th term = 15 th term = 41 Hence the upper quartile is 41. (iv)Interquartile range = Q3-Q1 = 41-27




= 14 Hence the Interquartile range is 14. 11.From the following frequency distribution, find : (i) the median (ii) lower quartile (iii) upper quartile (iv) inter quartile range Variate 15 18 20 22 25 27 30

Frequency 4 6 8 9 7 8 6

Solution: We write the variates in cumulative frequency table. Variate Frequency (f) Cumulative frequency 15 4 4 18 6 10 20 8 18 22 9 27 25 7 34 27 8 42 30 6 48 (i) Here number of observations, n = 48 which is even. So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (48/2 th term + ((48/2)+1)th term) = ½ (24 th term + (24+1)th term) = ½ (24 th term + 25th term) = ½ (22+22) [Since all observations from 19th to 27th are 22] = ½ ×44 = 22 Hence the median is 22. (ii) Lower quartile, Q1 = (n/4) th term = (48)/4 = 12 th term = 20 Hence the lower quartile is 20. (iii)Upper quartile, Q3 = (3n/4) th term = (3×48/4) th term = (3×12)th term = 36 th term = 27 Hence the upper quartile is 27.




(iv)Interquartile range = Q3-Q1 = 27-20 = 7 Hence the Interquartile range is 7. 12. For the following frequency distribution, find : (i) the median (ii) lower quartile (iii) upper quartile Variate 25 31 34 40 45 48 50 60

Frequency 3 8 10 15 10 9 6 2

Solution: We write the variates in cumulative frequency table. Variate Frequency (f) Cumulative frequency 25 3 3 31 8 11 34 10 21 40 15 36 45 10 46 48 9 55 50 6 61 60 2 63 (i) Here number of observations, n = 63 which is odd. Median = ((n+1)/2)th term = (63+1)/2 = 64/2 = 32th term = 40 Hence the median is 40. (ii) Lower quartile, Q1 = ((n+1)/4) th term = (63+1)/4 = 64/4 = 16 th term = 34 Hence the lower quartile is 34. (iii)Upper quartile, Q3 = (3(n+1)/4) th term = (3×(63+1)/4) th term = (3×(64/4)) th term = (3×16) th term




= 48 th term = 48 Hence the upper quartile is 48.




Exercise 21.3 1.Find the mode of the following sets of numbers ; (i) 3, 2, 0, 1, 2, 3, 5, 3 (ii) 5, 7, 6, 8, 9, 0, 6, 8, 1, 8 (iii) 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 Solution: Mode is the number which appears most often in a set of numbers. (i)Given set is 3, 2, 0, 1, 2, 3, 5, 3. In this set, 3 occurs maximum number of times. Hence the mode is 3. (ii) Given set is 5, 7, 6, 8, 9, 0, 6, 8, 1, 8. In this set, 8 occurs maximum number of times. Hence the mode is 8. (iii) Given set is 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7. In this set, 5 occurs maximum number of times. Hence the mode is 5. 2. Calculate the mean, the median and the mode of the numbers : 3, 2, 6, 3, 3, 1, 1, 2 Solution: We arrange given data in ascending order 1, 1, 2, 2, 3, 3, 3, 6 Mean = Ʃxi/n = (1+1+2+2+3+3+3+6)/8 = 21/8 = 2.625 Hence the mean is 2.625. Here number of observations, n = 8 which is even. So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (8/2 th term + ((8/2)+1)th term) = ½ (4 th term + (4+1)th term) = ½ (4 th term + 5th term) = ½ (2+3) = ½ ×5 = 2.5 Hence the median is 2.5. In the given set, 3 occurs maximum number of times. Hence the mode is 3. 3. Find the mean, median and mode of the following distribution : 8, 10, 7, 6, 10, 11, 6, 13, 10 Solution: We arrange given data in ascending order 6, 6, 7, 8, 10, 10, 10, 11, 13 Mean = Ʃxi/n = (6+6+7+8+10+10+10+11+13)/9




= 81/9 = 9 Hence the mean is 9. Here number of observations, n = 9 which is odd. Median = ((n+1)/2)th term = (9+1)/2 = 10/2 = 5th term = 10 Hence the median is 10. In the given set, 10 occurs maximum number of times. Hence the mode is 10. 4. Calculate the mean, the median and the mode of the following numbers : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2 Solution: We arrange given data in ascending order 1, 2, 3, 3, 3, 4, 5, 5, 6, 7 Mean = Ʃxi/n = (1+2+3+3+3+4+5+5+6+7)/10 = 39/10 = 3.9 Here number of observations, n = 10 which is even. So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (10/2 th term + ((10/2)+1)th term) = ½ (5 th term + (5+1)th term) = ½ (5 th term + 6th term) = ½ (3+4) = ½ ×7 = 3.5 Hence the median is 3.5. In the given set, 3 occurs maximum number of times. Hence the mode is 3. 5. The marks of 10 students of a class in an examination arranged in ascending order are as follows: 13, 35, 43, 46, x, x +4, 55, 61,71, 80 If the median marks is 48, find the value of x. Hence, find the mode of the given data. (2017) Solution: Given data in ascending order: 13, 35, 43, 46, x, x +4, 55, 61,71, 80 Given median = 48 Number of observations, n = 10 which is even. ∴ median = ½ ( n/2 th term + ((n/2)+1)th term) 48 = ½ (10/2 th term + ((10/2)+1)th term) 48 = ½ (5 th term + (5+1)th term) 48 = ½ (5 th term + 6th term) 48 = ½ (x+x+4)




48 = ½ ×(2x+4) 48 =x+2 ∴x = 48-2 = 46 ∴x+4 = 46+4 = 50 So the distribution becomes 13, 35, 43, 46, 46, 50, 55, 61,71, 80 Here 46 occurs maximum number of times. Hence the mode is 46. 6.A boy scored the following marks in various class tests during a term each test being marked out of 20: 15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16 (i) What are his modal marks ? (ii) What are his median marks ? (iii) What are his mean marks ? Solution: (i)We arrange given marks in ascending order 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19 16 appears maximum number of times. Hence his modal mark is 16. (ii)Here number of observations, n = 11 which is odd. So Median = ((n+1)/2)th term = (11+1)/2 = 12/2 = 6th term = 15 Hence the median is 15. (iii) Mean = Ʃxi/n = 7+ 10+12+12+14+ 15+16+16+16+ 17+ 19 = 154/11 = 14 Hence the mean is 14. 7. Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks : 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8 Solution: Given data is 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8 Number of observations, n = 16 Mean = Ʃxi/n = (0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8)/16 = 64/16 = 4 Hence the mean is 4. Here n = 16 which is even. So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (16/2 th term + ((16/2)+1)th term)




= ½ (8 th term + (8+1)th term) = ½ (8 th term + 9th term) = ½ (4+5) = 9/2 = 4.5 Hence the median is 4.5. Here 5 appears maximum number of times. Hence mode is 5. 8. Find the mode and median of the following frequency distribution : x 10 11 12 13 14 15 f 1 4 7 5 9 3 Solution: We write the data in cumulative frequency table. x Frequency (f) Cumulative frequency 10 1 1 11 4 5 12 7 12 13 5 17 14 9 26 15 3 29 Here number of observations, n = 29 which is odd. Median = ((n+1)/2)th term = (29+1)/2 = 30/2 = 15th term = 13 Hence the median is 13. Here the frequency corresponding to 14 is maximum. Hence the mode is 14. 9. The marks obtained by 30 students in a class assessment of 5 marks is given below: Marks 0 1 2 3 4 5 No. of students

1 3 6 10 5 5

Calculate the mean, median and mode of the above distribution. Solution: We write the data in cumulative frequency table. Marks x Frequency (f) Cumulative frequency fx 0 1 1 0 1 3 4 3 2 6 10 12




3 10 20 30 4 5 25 20 5 5 30 25 Total Ʃf = 30 Ʃfx = 90 Mean = Ʃfx/Ʃf = 90/30 = 3 Hence the mean is 3. Here number of observations, n = 30 which is even. So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (30/2 th term + ((30/2)+1)th term) = ½ (15 th term + (15+1)th term) = ½ (15 th term + 16th term) = ½ (3+3) = 6/2 = 3 Hence the median is 3. Here the mark 3 occurs maximum number of times. Hence the mode is 3. 10. The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution. Marks obtained

5 6 7 8 9 10

No. of students

3 9 6 4 2 1

Solution: We write the marks in cumulative frequency table. Marks x Frequency (f) fx Cumulative frequency 5 3 15 3 6 9 54 12 7 6 42 18 8 4 32 22 9 2 18 24 10 1 10 25 Total Ʃf = 25 Ʃfx = 171 Mean = Ʃfx/Ʃf = 171/25 = 6.84 Hence the mean is 6.84. Here number of observation, n = 25 which is odd. Median = ((n+1)/2)th term




= (25+1)/2 = 26/2 = 13th term = 7 Hence the median is 7. Here the frequency corresponding to 6 is maximum. Hence the mode is 6. 11. At a shooting competition, the scores of a competitor were as given below : Score 0 1 2 3 4 5 No. of shots

0 3 6 4 7 5

(i) What was his modal score ? (ii) What was his median score ? (iii) What was his total score ? (iv) What was his mean ? Solution: We write the marks in cumulative frequency table. Score x No. of shots (f) fx Cumulative frequency 0 0 0 0 1 3 3 3 2 6 12 9 3 4 12 13 4 7 28 20 5 5 25 25 Total Ʃf = 25 Ʃfx = 80 (i) Here the frequency corresponding to 4 is maximum. 4 occurs 7 times. Hence his modal score is 4. (ii)Here number of observation, n = 25 which is odd. Median = ((n+1)/2)th term = (25+1)/2 = 26/2 = 13th term = 3 Hence his median score is 3. (iii)Total score = Ʃfx = 80 Hence his total score is 80. (iv)Mean = Ʃfx/Ʃf = 80/25 = 3.2




Hence his mean score is 3.2. 12. (i) Using step-deviation method, calculate the mean marks of the following distribution. (ii) State the modal class. Class interval

50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90

Frequency 5 20 10 10 9 6 12 8 Solution: (i) Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 67.5 Class size (h) = 5

Class Interval Frequency (fi) Class mark (xi) di = xi - A ui = di/h fiui 50-55 5 52.5 -15 -3 -15 55-60 20 57.5 -10 -2 -40 60-65 10 62.5 5 -1 -10 65-70 10 67.5 10 0 0 70-75 9 72.5 15 1 9 75-80 6 77.5 20 2 12 80-85 12 82.5 25 3 36 85-90 8 87.5 30 4 32

Total ∑fi = 80 ∑fiui = 24


= 67.5+5(24/80) = 67.5+5×0.3 = 67.5+1.5 = 69 Hence the mean of the distribution is 69. (ii) Modal class is the class with highest frequency. Here the modal class is 55-60. 13. The following table gives the weekly wages (in Rs.) of workers in a factory : Weekly wages (in Rs)

50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90

No. of workers

5 20 10 10 9 6 12 8

Calculate: (i) The mean. (ii) the modal class (iii) the number of workers getting weekly wages below Rs. 80. (iv) the number of workers getting Rs. 65 or more but less than Rs. 85 as weekly wages.




Solution: We write the given data in cumulative frequency table. Class Interval Frequency (fi) Class mark (xi) Cumulative frequency fixi

50-55 5 52.5 5 262.5 55-60 20 57.5 25 1150 60-65 10 62.5 35 625 65-70 10 67.5 45 675 70-75 9 72.5 54 652.5 75-80 6 77.5 60 465 80-85 12 82.5 72 990 85-90 8 87.5 80 700

Total ∑fi = 80 ∑fixi = 5520

(i) Mean = ∑fixi /∑fi = 5520/80 = 69 Hence the mean is 69. (ii) Modal class is the class with highest frequency. Here the modal class is 55-60. (iii) The number of workers getting weekly wages below Rs. 80 is 60. [Check the cumulative frequency column and class interval column. 60 workers get below Rs. 80] (iv) The number of workers getting Rs. 65 or more but less than Rs. 85 as weekly wages = 72-35 = 37 [Check the cumulative frequency column and class interval column. ]




Exercise 21.4 1. Draw a histogram for the following frequency distribution and find the mode from the graph : Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5 Solution: Construct histogram using given data. Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5 Represent class on X-axis and frequency on Y-axis.

In the highest rectangle, draw two straight lines AC and BD. P is the point of intersection. Draw a vertical line through P to meet the X-axis at M. The abscissa of M is 14. Hence the mode is 14. 2. Find the modal height of the following distribution by drawing a histogram :




Height (in cm)

140-150 150-160 160-170 170-180 180-190

No. of students

7 6 4 10 2

Solution: Construct histogram using given data. Height (in cm)

140-150 150-160 160-170 170-180 180-190

No. of students

7 6 4 10 2

Represent height on X-axis and number of students on Y-axis. Take scale: X axis : 2 cm = 10 (class interval) Y axis : 1 cm = 1 (frequency)

In the highest rectangle, draw two straight lines AC and BD. P is the point of intersection. Draw a vertical line through P to meet the X-axis at M. The abscissa of M is 174. Hence the mode is 174. 3. A Mathematics aptitude test of 50 students was recorded as follows : Marks 50-60 60-70 70-80 80-90 90-100 No. of 4 8 14 19 5




students Draw a histogram for the above data using a graph paper and locate the mode. (2011) Solution: Construct histogram using given data. Marks 50-60 60-70 70-80 80-90 90-100 No. of students

4 8 14 19 5

Represent marks on X-axis and number of students on Y-axis. Take scale: X axis : 2 cm = 10 (class interval) Y axis : 1 cm = 1 (frequency)

In the highest rectangle, draw two straight lines AC and BD. P is the point of intersection. Draw a vertical line through P to meet the X-axis at M. The abscissa of M is 82.5. Hence the mode is 82.5. 4. Draw a histogram and estimate the mode for the following frequency distribution : Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 8 10 5 4 3




Solution: Construct histogram using given data. Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 8 10 5 4 3 Represent classes on X-axis and frequency on Y-axis. Take scale: X axis : 2 cm = 10 (class interval) Y axis : 1 cm = 1 (frequency)

In the highest rectangle, draw two straight lines AC and BD. P is the point of intersection. Draw a vertical line through P to meet the X-axis at M. The abscissa of M is 23. Hence the mode is 23. 5. IQ of 50 students was recorded as follows. IQ score 80-90 90-100 100-110 110-120 120-130 130-140 No. of students

6 9 16 13 4 2

Draw a histogram for the above data and estimate the mode. Solution: Construct histogram using given data.




Represent classes on X-axis and frequency on Y-axis. Take scale: X axis : 1 cm = 10 (class interval) Y axis : 1 cm = 1 (frequency)

In the highest rectangle, draw two straight lines AB and CD. M is the point of intersection. Draw a vertical line through M to meet the X-axis at L. The abscissa of L is 107. Hence the mode is 107. 6. Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below: Pocket expenses (in Rs) Number of students

(Frequency ) 0-5 10 5-10 14 10-15 28 15-20 42 20-25 50 25-30 30 30-35 14




35-40 12 Draw a histogram representing the above distribution and estimate the mode from the graph. Solution: Construct histogram using given data. Represent classes on X-axis and frequency on Y-axis. Take scale: X axis : 2 cm = 5 (class interval) Y axis : 1 cm = 5 (frequency)

In the highest rectangle, draw two straight lines AC and BD. P is the point of intersection. Draw a vertical line through P to meet the X-axis at M. The abscissa of M is 21. Hence the mode is 21. 7. Draw a histogram for the following distribution : Wt. in kg 40-44 45-49 50-54 55-59 60-64 65-69 No. of students

2 8 12 10 6 4

Hence estimate the modal weight. Solution: The given distribution is not continuous. Adjustment factor = (45-44)/2 = ½ = 0.5




We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit. So the new table in continuous form is given below. Weight in kg Number of students

(Frequency ) 39.5-44.5 2 44.5-49.5 8 49.5-54.5 12 54.5-59.5 10 59.5-64.5 6 64.5-69.5 4 Construct histogram using given data. Represent weight on X-axis and no. of students on Y-axis. Take scale: X axis : 2 cm = 5 (class interval) Y axis : 1 cm = 1 (frequency)

In the highest rectangle, draw two straight lines AC and BD. P is the point of intersection. Draw a vertical line through P to meet the X-axis at M. The abscissa of M is 52.75. Hence the mode is 52.75.




8. Find the mode of the following distribution by drawing a histogram Mid value 12 18 24 30 36 42 48 Frequency 20 12 8 24 16 8 12 Also state the modal class. Solution: Mid value Frequency 12 20 18 12 24 8 30 24 36 16 42 8 48 12 Here mid value and frequency is given. We can find the class size, h by subtracting second mid value from first mid value. ∴h = 18-12 = 6 So to find the lower limit of class interval, we subtract h/2 to the mid value. To find the upper limit of class interval, we add h/2 to the mid value. Here h/2 = 6/2 = 3 So lower limit = 12-3 = 9 Upper limit = 12+3 = 15 So the class interval is 9-15 Likewise we find the class interval of other values. Mid value Class interval Frequency

12 9-15 20 18 15-21 12 24 21-27 8 30 27-33 24 36 33-39 16 42 39-45 8 48 45-51 12 Construct histogram using given data. Take scale: X axis : 2 cm = 6 (class interval) Y axis : 1 cm = 2 (frequency)




In the highest rectangle, draw two straight lines AB and CD. M is the point of intersection. Draw a vertical line through M to meet the X-axis at L. The abscissa of L is 30.5. Hence the mode is 30.5. Modal class is the class with highest frequency. Hence the modal class is 27-33.




Exercise 21.5 1.Draw an ogive for the following frequency distribution: Height ( in cm )

150-160 160-170 170-180 180-190 190-200

No. of students

8 3 4 10 2

Solution: We write the given data in cumulative frequency table. Height in cm No of students Cumulative frequency 150-160 8 8 160-170 3 11 170-180 4 15 180-190 10 25 190-200 2 27 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis). Plot the points (160, 8), (170, 11), (180, 15), (190, 25) and (200, 27) on the graph. Join the points with the free hand. We get an ogive as shown:

2. Draw an ogive for the following data:




Class intervals

1-10 11-20 21-30 31-40 41-50 51-60

Frequency 3 5 8 7 6 2 Solution: The given distribution is not continuous. Adjustment factor = (11-10)/2 = ½ = 0.5 We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit. So the new table in continuous form is given below. We write the given data in cumulative frequency table. Class intervals frequency Cumulative frequency 0.5-10.5 3 3 10.5-20.5 5 8 20.5-30.5 8 16 30.5-40.5 7 23 40.5-50.5 6 29 50.5-60.5 2 31 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis). Plot the points (10.5, 3), (20.5, 8), (30.5, 16), (40.5, 23) , (50.5, 29) and (60.5, 31) on the graph. Join the points with the free hand. We get an ogive as shown:




3. Draw a cumulative frequency curve for the following data: Marks obtained

24-29 29-34 34-39 39-44 44-49 49-54 54-59

No. of students

1 2 5 6 4 3 2

Solution: We write the given data in cumulative frequency table. Marks obtained No of students Cumulative frequency 24-29 1 1




29-34 2 3 34-39 5 8 39-44 6 14 44-49 4 18 49-54 3 21 54-59 2 23 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis). Plot the points (29, 1), (34, 3), (39, 8), (44, 14), (49, 18) , (54, 21) and (59, 23) on the graph. Join the points with the free hand. We get an ogive as shown:




Exercise 21.6 1. The following table shows the distribution of the heights of a group of a factory workers. Height ( in cm )

150-155 155-160 160-165 165-170 170-175 175-180 180-185

No. of workers

6 12 18 20 13 8 6

(i) Determine the cumulative frequencies. (ii) Draw the cumulative frequency curve on a graph paper. Use 2 cm = 5 cm height on one axis and 2 cm = 10 workers on the other. (iii) From your graph, write down the median height in cm. Solution: (i) We write the given data in cumulative frequency table. Height in cm No of workers f Cumulative frequency 150-155 6 6 155-160 12 18 160-165 18 36 165-170 20 56 170-175 13 69 175-180 8 77 180-185 6 83 (ii)Plot the points (155, 6), (160, 18), (165, 36), (170, 56), (175, 69), (180, 77) and (185, 83) on the graph. Join the points with the free hand. We get an ogive as shown:




(iii)Here n = 83 , which is odd. So median = ((n+1)/2)th observation = ((83+1)/2)th observation = (84/2)th observation = 42th observation Take a point A(42) on Y-axis. From A, draw a horizontal line parallel to X-axis meeting the curve at B. From B draw a line perpendicular on the x-axis which meets it at C. ∴C is the median which is 166.5 cm. 2. Using the data given below construct the cumulative frequency table and draw the-Ogive. From the ogive determine the median.




Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 No. of students

3 8 12 14 10 6 5 2

Solution: We write the given data in cumulative frequency table. Marks No of students f Cumulative frequency c.f 0-10 3 3 10-20 8 11 20-30 12 23 30-40 14 37 40-50 10 47 50-60 6 53 60-70 5 58 70-80 2 60 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (10, 3), (20, 11), (30, 23), (40, 37), (50, 47), (60, 53), (70, 58) and (80, 60) on the graph. Join the points with the free hand. We get an ogive as shown:




Here number of observations, n = 60 which is even. So median = ( n/2) th term = (60/2 th term = 30 th term Mark a point A(30) on Y-axis. From A, draw a horizontal line parallel to X-axis meeting the curve at P. From P draw a line perpendicular on the x-axis which meets it at Q. ∴Q is the median . Q = 35 Hence the median is 35 . 3. Use graph paper for this question. The following table shows the weights in gm of a sample of 100




potatoes taken from a large consignment: Weight (gm)

50-60 60-70 70-80 80-90 90-100 100-110 110-120 120-130

Frequency 8 10 12 16 18 14 12 10 (i) Calculate the cumulative frequencies. (ii) Draw the cumulative frequency curve and from it determine the median weight of the potatoes. (1996) Solution: (i)We write the given data in cumulative frequency table. Marks frequency f Cumulative frequency c.f 50-60 8 8 60-70 10 18 70-80 12 30 80-90 16 46 90-100 18 64 100-110 14 78 110-120 12 90 120-130 10 100 (ii)To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (60, 8), (70, 18), (80, 30), (90, 46), (100, 64), (110, 78), (120, 90) and (130, 100) on the graph. Join the points with the free hand. We get an ogive as shown:




Here n = 100 which is even. So median = ( n/2 th term) = (100/2 th term) = (50 th term) Now mark a point A (50) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at P. From P, draw a perpendicular on x-axis meeting it at Q. Q is the median. Q = 93 gm. Hence the median is 93. 4. Attempt this question on graph paper.




Age( yrs) 5-15 15-25 25-35 35-45 45-55 55-65 65-75

No. of casualities due to accidents

6 10 15 13 24 8 7

(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years, on one axis and 2 cm = 10 casualties on the other. (ii) From your graph determine (1) the median and (2) the upper quartile Solution: (i)We write the given data in cumulative frequency table. Age (yrs) No of casualities due to accidents

f Cumulative frequency c.f

5-15 6 6 15-25 10 16 25-35 15 31 35-45 13 44 45-55 24 68 55-65 8 76 65-75 7 83 (ii)To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (15, 6), (25, 16), (35, 31), (45, 44), (55, 68), (65, 76) and (75, 83) on the graph. Join the points with the free hand. We get an ogive as shown:




(ii)(1). Here n = 83, which is odd. So median = (n+1)/2)th term = ((83+1)/2) = 84/2 = 42 Now mark a point A (42) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at P. From P, draw a perpendicular on x-axis meeting it at Q. Q is the median. Q = 43 Hence the median is 43.




(ii)(2). Upper quartile = (3(n+1)/4) = (3×(83+1)/4) = (3×(84)/4) = 63 Now mark a point B (63) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at L. From L, draw a perpendicular on x-axis meeting it at M. M = 52 Hence the upper quartile is 52. 5. The weight of 50 workers is given below: Weight in kg

50-60 60-70 70-80 80-90 90-100 100-110 110-120

No. of workers

4 7 11 14 6 5 3

Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis , and 2 cm = 5 workers along the other axis. Use a graph to estimate the following: (i) the upper and lower quartiles. (ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight. (2015) Solution: We write the given data in cumulative frequency table. Weight in kg No of workers f Cumulative frequency c.f 50-60 4 4 60-70 7 11 70-80 11 22 80-90 14 36 90-100 6 42 100-110 5 47 110-120 3 50 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50) on the graph. Join the points with the free hand. We get an ogive as shown:




(i)Here n = 50, which is even. Upper quartile = 3n/4 = 3×50/4 = 150/4 = 37.5 Now mark a point A (37.5) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at B. From B, draw a perpendicular on x-axis meeting it at C. C = 92.5 Hence the upper quartile is 92.5 kg. Lower quartile, Q1 = (n/4) th term = 50/4 = 12.5 Now mark a point D(12.5) on the Y-axis and from D draw a line parallel to X-axis meeting the curve at E. From




E, draw a perpendicular on x-axis meeting it at F. F = 72 Hence the lower quartile is 72 kg. (ii) Mark on the graph point P which is 95 kg on X axis. Through P draw a vertical line to meet the ogive at Q. Through Q, draw a horizontal line to meet y-axis at R. The ordinate of point R represents 40 workers on the y-axis . ∴The number of workers who are 95 kg and above = Total number of workers – number of workers of weight less than 95 kg = 50-40 = 10 6. The table shows the distribution of scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis) Scores 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of shooters

9 13 20 26 30 22 15 10 8 7

Use your graph to estimate the following: (i) The median. (ii) The interquartile range. (iii) The number of shooters who obtained a score of more than 85%. Solution: We write the given data in cumulative frequency table. Scores No of shooters f Cumulative frequency c.f 0-10 9 9 10-20 13 22 20-30 20 42 30-40 26 68 40-50 30 98 50-60 22 120 60-70 15 135 70-80 10 145 80-90 8 153 90-100 7 160 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153) and (100, 160) on the graph. Join the points with the free hand. We get an ogive as shown:




(i)Here n = 160, which is even. So median = n/2 = 80 Now mark a point A(80) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at B. From P, draw a perpendicular on x-axis meeting it at C. C is the median. C = 44 (ii) lower quartile, Q1 = (n/4)th term = 160/4 = 40 Now mark a point D(40) on the Y-axis and from that point draw a line parallel to X-axis meeting the curve at E. From E, draw a perpendicular on x-axis meeting it at F. F = 29




So Q1= 29 Upper quartile, Q3 = (3n/4)th term = 3×160/4 = 3×40 = 120 Mark a point P(120) on the Y-axis and from that point draw a line parallel to X-axis meeting the curve at Q. From Q, draw a perpendicular on x-axis meeting it at R. R = 60 So Q3 = 60 Inter quartile range = Q3- Q1

= 60-29 = 31 Hence the Inter quartile range is 31. (iii) Mark a point Z(85) on the X axis. From Z on X-axis, draw a perpendicular to it meeting the curve at Y. From Y, draw a line parallel to X-axis meeting Y-axis at X. X is the required point which is 150. Number of shooters getting more than 85% scores = Total number of shooters - number of shooters who got till 85% = 160-150 = 10. Hence the number of shooters getting more than 85% scores is 10. 7. The daily wages of 80 workers in a project are given below Wages in Rs 400-450 450-500 500-550 550-600 600-650 650-700 700-750

No. of workers

2 6 12 18 24 13 5

Use a graph paper to draw an ogive for the above distribution. ( a scale of 2 cm = Rs 50 on x- axis and 2 cm = 10 workers on y-axis). your ogive to estimate (i) the median wage of the workers. (ii) the lower quartile wage of the workers. (iii) the number of workers who earn more than Rs 625 daily. (2017) Solution: We write the given data in cumulative frequency table. Wages in Rs. No of workers f Cumulative frequency c.f 400-450 2 2 450-500 6 8 500-550 12 20 550-600 18 38 600-650 24 62 650-700 13 75 700-750 5 80 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (450, 9), (500, 22), (550, 42), (600, 68), (650, 98), (700, 120) and (750, 135) on the graph. Join the points with the free hand. We get an ogive as shown:




(i)Here n = 80. Median = (n/2)th term = 80/2 = 40th term Mark a point (40) on Y axis. Draw a line from that point parallel to X axis. Let it meet the curve at A. Draw a perpendicular from A to meet X axis at B. The value of B is 604. Hence the median is 604. (ii)Lower quartile, Q1 = (n/4)th term




= 80/4 = 20th term = 550 [from graph] (iii) Draw a vertical line through the point 625 on X axis. which meets the graph at point C. From C, draw a horizontal line which meets the y-axis at the mark of 50. Thus, number of workers that earn more Rs 625 daily = Total no. of workers - no. of workers who earn upto 625 = 80-50 = 30 8. Marks obtained by 200 students in an examination are given below marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of students

5 11 10 20 28 37 40 29 14 6

Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph, determine (i) The median marks. (ii) The number of students who failed if minimum marks required to pass is 40. (iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination. Solution: We write the given data in cumulative frequency table. Marks No of students f Cumulative frequency

c.f 0-10 5 5 10-20 11 16 20-30 10 26 30-40 20 46 40-50 28 74 50-60 37 111 60-70 40 151 70-80 29 180 80-90 14 194 90-100 6 200 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111) , (70, 151) (80, 180) (90, 194) and (100, 200) on the graph. Join the points with the free hand. We get an ogive as shown:




(i)Here n= 200 Median = (n/2)th term = 200/2 = 100th term = 57 [from graph] (ii)number of students failed if minimum marks required to pass is 40 = 44 [from graph] (iii)Number of students who got grade 1 = number of students who scored 85 and more = 200-188 = 12[From graph] 9. The monthly income of a group of 320 employees in a company is given below Monthly income No. of employees 6000-7000 20




7000-8000 45 8000-9000 65 9000-10000 95 10000-11000 60 11000-12000 30 12000-13000 5 Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine (i) the median wage. (ii) the number of employees whose income is below Rs. 8500. (iii) If the salary of a senior employee is above Rs. 11500, find the number of senior employees in the company. (iv) the upper quartile. Solution: We write the given data in cumulative frequency table. Monthly income No. of employees Cumulative frequency

c.f 6000-7000 20 20 7000-8000 45 65 8000-9000 65 130 9000-10000 95 225 10000-11000 60 285 11000-12000 30 315 12000-13000 5 320 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (7000, 20), (8000, 65), (9000, 130), (10000, 225), (11000, 285), (12000, 315) and (13000, 320) on the graph. Join the points with the free hand. We get an ogive as shown:




(i)here n = 320 Median = (n/2)th term = 320/2 = 160th term Mark the point A(160) on Y axis. Draw a line parallel to x axis from that point. Let it meet the curve at P. Draw a perpendicular from P to X axis which meets at M. M is the median. Here median is 9300. [from graph]




(ii)Mark the point B(8500) on X axis. Draw a line parallel to Y axis which meets curve at Q. From Q draw a line parallel to X axis which meets Y axis at N. N = 98 number of employees whose income is below 8500 = 98 (iii) Mark the point C(11500) on the x-axis. Draw a line perpendicular to x-axis meeting the curve at R. From R, draw a line parallel to x-axis meeting y-axis at L which is 300 No. of employees getting more than Rs. 11500 = 320-300 = 20 (iv)upper quartile = 3n/4 = 3×320/4 = 240 Mark the point T(240) on Y axis. From that point on y-axis, draw a line perpendicular on the x-axis which meets the curve at S. From S, draw a perpendicular on x-axis meeting it at U, which is 10250. Hence upper quartile is 10250. 10. Using a graph paper, draw an ogive for the following distribution which shows a record of the weight in kilograms of 200 students Weight 40-45 45-50 50-55 55-60 60-65 65-70 70-75 75-80

Frequency 5 17 22 45 51 31 20 9

Use your ogive to estimate the following: (i) The percentage of students weighing 55 kg or more. (ii) The weight above which the heaviest 30% of the students fall, (iii) The number of students who are : 1. under-weight and 2. over-weight, if 55.70 kg is considered as standard weight. Solution: We write the given data in cumulative frequency table. Weight Frequency Cumulative frequency

c.f 40-45 5 5 45-50 17 22 50-55 22 44 55-60 45 89 60-65 51 140 65-70 31 171 70-75 20 191 75-80 9 200 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200) on the graph. Join the points with the free hand. We get an ogive as shown:




(i)Total number of students = 200 The number of students weighing 55 kg or more = 200-44 = 156 [From the graph] Percentage = (156/200)×100 = 156/2 = 78%. (ii)30% of 200 = (30/100)×200 = 30×2




= 60 No of heaviest students = 31+20+9 = 60 60 students fall above 65 kg. (iii)If 55.70 kg is the standard weight, No. of students who are under weight = 47 [from graph] No. of students who are overweight = 200-47 = 153 11. The marks obtained by 100 students in a Mathematics test are given below Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of students

3 7 12 17 23 14 9 6 5 4

Draw an ogive on a graph sheet and from it determine the : (i) median (ii) lower quartile (iii) number of students who obtained more than 85% marks in the test. (iv) number of students who did not pass in the test if the pass percentage was 35. Solution: We write the given data in cumulative frequency table. Marks No. of students Cumulative frequency

c.f 0-10 3 3 10-20 7 10 20-30 12 22 30-40 17 39 40-50 23 62 50-60 14 76 60-70 9 85 70-80 6 91 80-90 5 96 90-100 4 100 Plot the points (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85), (80, 91), (90, 96) and (100, 100) on the graph. Join the points with the free hand. We get an ogive as shown:




(i)Here n = 100 Median = n/2 = 50 th term Mark a point A(50) on Y axis. Draw a line parallel to X axis from A. Let it meet the curve at B. From B draw a perpendicular which meets X axis at C. The point C is 45. Hence median is 45. (ii)Lower quartile = n/4




= 100/4 = 25th term Mark a point P (25) on Y axis. Draw a line parallel to X axis from that point. Let it meet the curve at Q. From that point draw a perpendicular which meets X axis at R. The point R is 32. Hence lower quartile is 32. (iii)no. of students who obtained more than 85% = 100-94 = 6 [from graph] (iv)No of students who failed if 35% is the pass percentage = 25 [from graph] 12. The marks obtained by 120 students in a Mathematics test are-given below Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of students

5 9 16 22 26 18 11 6 4 3

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for ogive to estimate the following: (i) the median (ii) the lower quartile (iii) the number of students who obtained more than 75% marks in the test. (iv) the number of students who did not pass in the test if the pass percentage was 40. (2002) Solution: We write the given data in cumulative frequency table. Marks No. of students Cumulative frequency

c.f 0-10 5 5 10-20 9 14 20-30 16 30 30-40 22 52 40-50 26 78 50-60 18 96 60-70 11 107 70-80 6 113 80-90 4 117 90-100 3 120 Plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117) and (100, 120) on the graph. Join the points with the free hand. We get an ogive as shown:




(i)Here n = 120 Median = (n/2)th term = 120/2 = 60th term Mark point A(60) on Y axis. Draw a line parallel to X axis from A.




Let it meet the curve at B. Draw a straight line from B to X axis which meets at C. C = 50 Hence median is 50. (ii)Lower quartile = (n/4)th term = 120/4 = 30th term Mark a point P (30) on Y axis. Draw a line parallel to X axis from that point. Let it meet the curve at Q. From that point draw a perpendicular which meets X axis at R. The point R is 30. Hence lower quartile is 30. (iii)Mark a point U(75) on X axis. Draw a line parallel to Y axis which meets curve at T. From T, draw a line parallel to X axis to meet Y axis at S. S = 110 No. of students who obtained more than 75% = 120-110 = 10 (iv) Mark a point Z(40) on X axis. Draw a line parallel to Y axis which meets curve at Y. From Y, draw a line parallel to X axis to meet Y axis at X. X = 52 No of students who failed if 40% is the pass percentage is 52. 13. The following distribution represents the height of 160 students of a school. Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180

No. of students

12 20 30 38 24 16 12 8

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine : (i)The median height. (ii)The inter quartile range. (iii) The number of students whose height is above 172 cm. Solution: We write the given data in cumulative frequency table. Height No. of students Cumulative frequency

c.f 140-145 12 12 145-150 20 32 150-155 30 62 155-160 38 100 160-165 24 124 165-170 16 140 170-175 12 152 175-180 8 160




Plot the points (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152), and (180, 160) on the graph. Join the points with the free hand. We get an ogive as shown:

(i)Here n = 160 Median = (n/2)th term = 160/2 = 80th term Mark point A(80) on Y axis. Draw a line parallel to X axis from A. Let it meet the curve at B. Draw a straight line from B to X axis which meets at C. C = 157.5 Hence median is 157.5. (ii) Lower quartile , Q1 = (n/4)th term = 160/4 = 40th term Proceeding in the same way mentioned in (i), we get lower quartile = 152 [Point R]




Upper quartile, Q3 = 3n/4 = 3×160/4 = 3×40 = 120th term Proceeding in the same way mentioned in (i), we get upper quartile = 164 [Point Z] Interquartile range = Q3-Q1 = 164-152 = 12 (iii)Mark a point O(172 ) on X axis. Draw a line parallel to Y axis from O. Let it meet the curve at N. Draw a straight line from N to Y axis which meets at M. M = 144 Hence number of students whose height is more than 172 cm is 160-144 = 16 14. 100 pupils in a school have heights as tabulated below Height in cm

121-130 131-140 141-150 151-160 161-170 171-180

No. of pupils

12 16 30 20 14 8

Draw the ogive for the above data and from it determine the median (use graph paper). Solution: We write the given data in cumulative frequency table (in continuous distribution): Height No. of students Cumulative frequency

c.f 120.5-130.5 12 12 130.5-140.5 16 28 140.5-150.5 30 58 150.5-160.5 20 78 160.5-170.5 14 92 170.5-180.5 8 100 Plot the points (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92) and (180.5, 100) on the graph. Join the points with the free hand. We get an ogive as shown:




Here n = 100 Median = (n/2)th term = 100/2 = 50th term Mark point A(50) on Y axis. Draw a line parallel to X axis from A. Let it meet the curve at P. Draw a straight line from P to X axis which meets at Q. Q = 147.5 Hence median is 147.5.




Chapter Test 1. Arun scored 36 marks in English, 44 marks in Civics, 75 marks in Mathematics and x marks in Science. If he has scored an average of 50 marks, find x. Solution: Marks scored in English = 36 Marks scored in Civics = 44 Marks scored in Mathematics = 75 Marks scored in Science = x No. of subjects = 4 Average marks = sum of marks / No. of subjects = 50 [Given] ∴(36+44+75+x)/4 = 50 ∴155+x = 4×50 ∴155+x = 200 ∴x = 200-155 ∴x = 45 Hence the value of x is 45. 2. The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers. Solution: Given the mean of 20 numbers = 18 ∴Sum of numbers = 18×20 = 360 If 3 is added to each of first 10 numbers, then new sum = (3×10)+360 = 30+360 = 390 ∴New mean = 390/20 = 19.5 Hence the mean of new set of 20 numbers is 19.5. 3. The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for computation of mean. Find the correct mean. Solution: Average height of 30 students = 150 cm So sum of height = 150×30 = 4500 Difference between correct value and wrong value = 165-135 = 30 So actual sum = 4500+30 = 4530 So actual mean = 4530/30 = 31 Hence the correct mean 31. 4. There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and average weight of the girls is 40 kg. Find the average weight of boys. Solution: Total number of students = 50 No. of boys = 40




∴No. of girls = 50-40 = 10 Average weight of 50 students = 44 kg So sum of weight = 44×50 = 2200 kg Average weight of girls = 40 kg So sum of weight of girls = 40×10 = 400 kg ∴Total weight of boys = 2200-400 = 1800 kg ∴Average weight of boys = 1800/40 = 45 kg Hence the average weight of boys is 45 kg. 5. The contents of 50 boxes of matches were counted giving the following results. No. of matches

41 42 43 44 45 46

No. of boxes

5 8 13 12 7 5

Calculate the mean number of matches per box. Solution: No. o f matches x No. of boxes f fx 41 5 205 42 8 336 43 13 559 44 12 528 45 7 315 46 5 230 Total Ʃf = 50 Ʃfx =2173 Mean = Ʃfx/Ʃf = 2173/50 = 43.46 Hence the mean is 43.46. 6. The heights of 50 children were measured (correct to the nearest cm) giving the following results Height (in cm)

65 66 67 68 69 70 71 72 73

No. of children

1 4 5 7 11 10 6 4 2

Calculate the mean height for this distribution correct to one place of decimal. Solution: Height x No. of children f fx 65 1 65 66 4 264 67 5 335 68 7 476 69 11 759




70 10 700 71 6 426 72 4 288 73 2 146 Total Ʃf = 50 Ʃfx = 3459 Mean = Ʃfx/Ʃf = 3459/50 = 69.18 = 69.2 [corrected to one decimal place] Hence the mean is 69.2. 7. Find the value of p for the following distribution whose mean is 20.6. Variate (xi) 10 15 20 25 35

Frequency (fi)

3 10 p 7 5

Solution: Variate (xi) Frequency (fi) fx 10 3 30 15 10 150 20 p 20p 25 7 175 35 5 175 Total Ʃfi = 25+p Ʃfi xi = 530+20p Mean = Ʃfx/Ʃf ∴20.6 = (530+20p )/(25+p) [Given mean = 20.6] ∴20.6(25+p) = (530+20p) ∴515 + 20.6p = 530+20p 20.6p-20p = 530-515 0.6p = 15 p = 15/0.6 p = 25 Hence the value of p is 25. 8. Find the value of p if the mean of the following distribution is 18. Variate (xi) 13 15 17 19 20+p 23

Frequency (fi)

8 2 3 4 5p 6

Solution: Variate (xi) Frequency (fi) fi xi




13 8 104 15 2 30 17 3 51 19 4 76 20+p 5p 5p2+100p 23 6 138 Total Ʃfi = 23+5p Ʃfi xi = 399+5p2+100p Mean = Ʃfi xi / Ʃfi ∴18 = (399+5p2+100p)/( 23+5p) [Given mean = 18] ∴18(23+5p) = 399+5p2+100p ∴414 + 90p = 399+5p2+100p ∴5p2+100p-90p+399-414 = 0 ∴5p2+10p-15 = 0 Dividing by 5, we get ∴p2+2p-3 = 0 ∴(p-1)(p+3) = 0 ∴p-1 = 0 or p+3 = 0 p = 1 or p = -3 p cannot be negative. So p = 1 Hence the value of p is 1. 9. Find the mean age in years from the frequency distribution given below: Age in years 25-29 30-34 35-39 40-44 45-49 50-54 55-59

No. of persons

4 14 22 16 6 5 3

Solution: The given distribution is not continuous. Adjustment factor = (30-29)/2 = ½ = 0.5 We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit. So the new table in continuous form is given below. Age Mid value xi No. of persons fi fixi 24.5-29.5 27 4 108 29.5-34.5 32 14 448 34.5-39.5 37 22 814 39.5-44.5 42 16 672 44.5-49.5 47 6 282 49.5-54.5 52 5 260 54.5-59.5 57 3 171 Total Ʃfi = 70 Ʃ fixi = 2755 Mean = Ʃfi xi / Ʃfi




= 2755/70 = 39.357 = 39.36 Hence the mean age is 39.36 years. 10. Calculate the Arithmetic mean, correct to one decimal place, for the following frequency Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

Students 2 4 5 16 20 10 6 8 4

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Marks Students fi Class mark xi fixi 10-20 2 15 30 20-30 4 25 100 30-40 5 35 175 40-50 16 45 720 50-60 20 55 1100 60-70 10 65 650 70-80 6 75 450 80-90 8 85 680 90-100 4 95 380 Total Ʃfi = 75 Ʃfixi = 4285 Mean = Ʃfi xi / Ʃfi = 4285/75 = 57.133 = 57.1 Hence the mean is 57.1. 11. The mean of the following frequency distribution is 62.8. Find the value of p. Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 5 8 p 12 7 8 Solution: Class Frequency fi Class mark xi fixi 0-20 5 10 50 20-40 8 30 240 40-60 p 50 50p 60-80 12 70 840 80-100 7 90 630 100-120 8 110 880 Total Ʃfi = 40+p Ʃfixi = 2640+50p Mean = Ʃfi xi / Ʃfi 62.8 = (2640+50p)/( 40+p) [Given mean = 62.8]




62.8(40+p) = 2640+50p 2512+62.8p = 2640+50p 62.8p-50p = 2640-2512 12.8p = 128 p = 128/12.8 ∴p = 10 Hence the value of p is 10. 12. The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188. Expenditure in Rs

140-160 160-180 180-200 200-220 220-240

No. of families

5 25 f1 f2 5

Solution: Given mean = 188 Class Frequency fi Class mark xi fixi 140-160 5 150 750 160-180 25 170 4250 180-200 f1 190 190f1

200-220 f2 210 210f2 220-240 5 230 1150 Total Ʃfi = 35+f1+f2 = 100 Ʃfixi = 6150+190f1+210f2

Given no. of families = 100 So 35+f1+f2 = 100 ∴ f1+f2 = 100-35 = 65 ∴ f1 = 65-f2 ..(i) Mean = Ʃfi xi / Ʃfi 188 = (6150+190f1+210f2)/100 [Given mean = 188] 188(100) = 6150+190f1+210f2

18800 = 6150+190f1+210f2

18800-6150 = 190f1+210f2

12650 = 190f1+210f2 ..(ii) Substitute (i) in (ii) 12650 = 190(65-f2)+210f2 12650 = 12350-190f2+210f2 12650-12350 = -190f2+210f2 300 = 20f2 ∴f2 = 300/20 = 15 Put f2 in (i) ∴ f1 = 65-15 ∴ f1 = 50 Hence the value of f1 and f2 is 50 and 15 respectively.

13. The measures of the diameter of the heads of 150 screw is given in the following table. If the mean




diameter of the heads of the screws is 51.2 mm, find the values of p and q . Diameter in mm

32-36 37-41 42-46 47-51 52-56 57-61 62-66

No. of screws

15 17 p 25 q 20 30

Solution: Given mean = 51.2 mm The given distribution is not continuous. Adjustment factor = (37-36)/2 = ½ = 0.5 We subtract 0.5 from lower limit of the class interval and add 0.5 to upper limit. So the new table in continuous form is given below. Diameter in mm Mid value xi No. of screws fi fixi 31.5-36.5 34 15 510 36.5-41.5 39 17 663 41.5-46.5 44 p 44p 46.5-51.5 49 25 1225 51.5-56.5 54 q 54q 56.5-61.5 59 20 1180 61.5-66.5 64 30 1920 Total Ʃfi = 107+p+q Ʃ fixi = 5498+44p+54q Given No. of screws = 150 ∴107+p+q = 150 p = 150-107-q p = 43-q ..(i) Mean = Ʃfi xi / Ʃfi ∴51.2 = (5498+44p+54q)/150 51.2×150 = 5498+44p+54q 7680 = 5498+44(43-q)+54q 7680 = 5498+1892-44q+54q 7680 -5498-1892 = -44q+54q 290 = 10q ∴q = 290/10 = 29 Put q in (i) p = 43-29 ∴p = 14 Hence the value of p and q is 14 and 29 respectively. 14. The median of the following numbers, arranged in ascending order is 25. Find x. 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46 Solution: Here n = 10, which is even Median = 25 So Median = ½ ( n/2 th term + ((n/2)+1)th term) ∴25 = ½ (( 10/2 )th term + (10/2)+1)th term) ∴25 = ½ (( 5 )th term + (6)th term)




∴25 = ½ (x+2 + x+4) ∴25 = ½ (2x+6) ∴2x+6 = 25×2 ∴2x+6 = 50 ∴2x = 50-6 = 44 ⇒x = 44/2 = 22 Hence the value of x is 22. 15. If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x. Solution: Arranging numbers in ascending order 3,4,5,x,8,9,11 Here n = 7 which is odd Given median = 6 So Median =( (n+1)/2) th term ∴6 = ((7+1)/2)th term ∴6 = ((8/2)th term ∴6 = 4th term ∴6 = x Hence the value of x is 6. 16. Find the median of: 17, 26, 60, 45, 33, 32, 29, 34, 56 . If 26 is replaced by 62, find the new median. Solution: Arranging numbers in ascending order 17,26,29,32,33,34,45,56,60 Here n = 9 which is odd. So Median =( (n+1)/2) th term ∴Median = ((9+1)/2)th term ∴Median = (10/2)th term ∴Median = 5th term ∴Median = 33 Hence median is 33. If 26 is replaced by 62, new set of numbers in ascending order is shown below. 17,29,32,33,34,45,56,60,62 Here n = 9 which is odd. So Median =( (n+1)/2) th term ∴Median = ((9+1)/2)th term ∴Median = (10/2)th term ∴Median = 5th term ∴Median = 34 Hence median is 34. 17. The marks scored by 16 students in a class test are : 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12 Find (i) the median




(ii) lower quartile (iii) upper quartile Solution: Arranging data in ascending order 1,3,5,6,8,9,10,12,13,15,17,18,20,21,21,23 Here n = 16 which is even (i) So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (16/2 th term + ((16/2)+1)th term) = ½ (8 th term + (8+1)th term) = ½ (8 th term + 9th term) = ½ (12+13) = ½ ×25 = 12.5 Hence the median is 12.5. (ii) Lower quartile, Q1 = (n/4) th term = (16)/4 = 4 th term = 6 Hence the lower quartile is 6. (iii)Upper quartile, Q3 = (3n/4) th term = (3×16/4) th term = (3×4)th term = 12 th term = 18 Hence the upper quartile is 18. 18. Find the median and mode for the set of numbers : 2, 2, 3, 5, 5, 5, 6, 8, 9 Solution: Here n = 9 which is odd. ∴Median = ((n+1)/2)th term ∴Median = ((9+1)/2)th term ∴Median = (10/2)th term ∴Median = 5 th term ∴Median = 5 Mode is the number which appears most often in a set of numbers. Here 5 occurs maximum number of times. So mode is 5. 19. Calculate the mean, the median and the mode of the following distribution. Age in years

12 13 14 15 16 17 18

No. of students

2 3 5 6 4 3 2

Solution:




Age in years xi No. of students fi Cumulative frequency fixi 12 2 2 24 13 3 5 39 14 5 10 70 15 6 16 90 16 4 20 64 17 3 23 51 18 2 25 36 Total Ʃfi = 25 Ʃ fixi = 374 Mean = Ʃ fixi/ Ʃfi = 374/25 = 14.96 Hence the mean is 14.96. Here n = 25 which is odd. ∴Median = ((n+1)/2)th term ∴Median = ((25+1)/2)th term ∴Median = (26/2)th term ∴Median = 13 th term ∴Median = 15 Hence the median is 15. Here 15 occurs maximum number of times. i.e., 6 times. Hence the mode is 15. 20. The daily wages of 30 employees in an establishment are distributed as follows: Daily wages in Rs

0-10 10-20 20-30 30-40 40-50 50-60

No. of employees

1 8 10 5 4 2

Estimate the modal daily wages for this distribution by a graphical method. Solution: Daily wages in Rs. No. of employees 0-10 1 10-20 8 20-30 10 30-40 5 40-50 4 50-60 2 Taking daily wages on x-axis and No. of employees on the y-axis and draw a histogram as shown below.




Join AB and CD intersecting each other at M. From M draw ML perpendicular to x-axis. L is the mode Here Mode = Rs 23 Hence the mode is Rs. 23. 21. Using the data given below, construct the cumulative frequency table and draw the ogive. From the ogive, estimate : (i) the median (ii) the inter quartile range. Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Frequency 3 8 12 14 10 6 5 2 Also state the median class Solution: Arranging the data in cumulative frequency table. Marks Frequency Cumulative frequency 0-10 3 3 10-20 8 11 20-30 12 23 30-40 14 37 40-50 10 47




50-60 6 53 60-70 5 58 70-80 2 60 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis). Plot the points (10, 3), (20, 11), (30, 23), (40, 37), (50, 47), (60, 53), (70, 58) and (80, 60) on the graph. Join the points with the free hand. We get an ogive as shown:

(i)Here number of observations, n = 60 which is even. So median = ( n/2) th term = (60/2) th term = 30 th term Mark a point A(30) on Y-axis. From A, draw a horizontal line parallel to X-axis meeting the curve at P. From P draw a line perpendicular to the x-axis which meets it at Q. ∴Q is the median . Q = 35 Hence the median is 35 . (ii) Lower quartile, Q1 = n/4 = 60/4 = 15th term Upper quartile, Q3 = 3n/4 = 3×60/4 = 45th term Mark a point B(15) and C(45) on Y-axis. From B and C, draw a horizontal line parallel to X-axis meeting the curve at L and M respectively. From L and M, draw lines perpendicular to the x-axis which meets it at E and F




respectively. E is the lower quartile . E = 22.3 F is the upper quartile. F = 47 Inter quartile range = Q3-Q1 = 47-22.3 = 24.7 Hence interquartile range is 24.7. 22. Draw a cumulative frequency curve for the following data : Marks obtained

0-10 10-20 20-30 30-40 40-50

No. of students

8 10 22 40 20

Hence determine: (i) the median (ii) the pass marks if 85% of the students pass. (iii) the marks which 45% of the students exceed. Solution: Arranging the data in cumulative frequency table. Marks obtained No. of students f Cumulative frequency 0-10 8 8 10-20 10 18 20-30 22 40 30-40 40 80 40-50 20 100 To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis). Plot the points (10, 8), (20, 18), (30, 40), (40, 80), and(50,100) on the graph. Join the points with the free hand. We get an ogive as shown:




(i) Here number of observations, n = 100 which is even. So median = ( n/2) th term = (100/2) th term = 50 th term Mark a point A(50) on Y-axis. From A, draw a horizontal line parallel to X-axis meeting the curve at P. From P, draw a line perpendicular to the x-axis which meets it at Q.




∴Q is the median . Q = 32.5 Hence the median is 32.5 . (ii)Total number of students = 100 85% of 100 = 85 Remaining number of students = 100-85 = 15 Mark a point B(15) on Y axis. From B, draw a horizontal line parallel to X-axis meeting the curve at L. From L, draw a line perpendicular to the x-axis which meets it at M. Here M = 18 The pass marks will be 18 if 85% of students passed. (iii) Total number of students = 100 45% of 100 = 45 Remaining number of students = 100-45 = 55 Mark a point C(55) on Y axis. From C, draw a horizontal line parallel to X-axis meeting the curve at E. From E, draw a line perpendicular to the x-axis which meets it at F. Here F = 34 Hence marks which 45% of students exceeds is 34 marks.



ml aggarwal solutions class 10 maths chapter 21: measures

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